Two circles with equal radii are intersecting at the points $(0, 1)$ and $(0, -1)$. The tangent at the point $(0, 1)$ to one of the circle. Then the distance between the centres of these circles is:

The two intersecting circles share a common chord connecting the points (0, 1) and (0, -1). The midpoint of this common chord is the origin (0, 0), and the length of the chord is 2. Because the two circles have equal radii, their centers lie on the perpendicular bisector of the common chord. Since the common chord is vertical, the perpendicular bisector is the horizontal x-axis (y = 0). Let the centers be (-a, 0) and (a, 0) for some positive distance a. The distance from either center to an intersection point, for instance (0, 1), gives the radius squared as a^2 + 1^2. The question states that the tangent at the point (0, 1) to one of the circles passes through the center of the other circle, meaning the radius to (0,1) is perpendicular to the tangent line connecting the centers. We verify this implies the triangle formed by the two centers and the point (0,1) is right-angled at (0,1). By symmetry, the distance from (0,1) to both (-a,0) and (a,0) is the same, making the distance between the two centers exactly twice the horizontal distance a. If the tangent line from (0,1) to the other center has a certain slope, the radius to its own center is perpendicular to it. Let the radius to center C1(-a,0) have slope (0-1)/(-a-0) = 1/a. The line from (0,1) to the other center C2(a,0) has slope (0-1)/(a-0) = -1/a. For the tangent to be perpendicular to the radius, their slopes must multiply to -1, and here (1/a) * (-1/a) = -1/a^2, which implies a^2 = 1, so a = 1. The total distance between the two centers is 2*a. The result is 2.

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